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Failure Rate Modeling

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Reliability Physics and Engineering

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Abstract

For a collection of devices, it is critically important to be able to understand the expected failure rate for the devices. For the supplier of such devices, the expected failure rate will be an important indicator of future warranty liability. For the customer, the expected failure rate will be an important indicator of future satisfaction. For mission-critical applications, it is of paramount importance for one to know that the expected failure rate will be extremely low.

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Notes

  1. 1.

    The use of the expression mission critical came into vogue for space applications. In space applications, device repair or replacement is very difficult, if not impossible. Therefore, it is imperative that such devices have extremely low failure rates. However, today, life-support implantable devices are widely used. If one of these devices is part of your life-support system, there is little doubt that you would describe this as a mission-critical application.

  2. 2.

    The identity \( 1=t/{\int}_0^t\mathrm{d}t \) is used in Eq. (8.4).

  3. 3.

    In order to be consistent with Eq. (8.1), the true unit of failure rate λ must be in reciprocal time. Often the pseudo units (failures and devices) are introduced for emphasis and to facilitate a little bookkeeping. However, the true units of the FIT are: 1 FIT = 10−9/h.

  4. 4.

    A common-experience analogy is perhaps useful—the instantaneous speed that you drive is usually far more important than your average speed. Speeding tickets are normally issued based on instantaneous speed, not average speed!

  5. 5.

    This is why the EFR region is also referred to as infant mortality.

Bibliography

  • McPherson, J: Accelerated Testing. In: Electronic Materials Handbook, Vol. 1 Packaging, ASM International, 887 (1989).

    Google Scholar 

  • Miller, I. and J. Freund: Probability and Statistics for Engineers, 2nd. Ed. Prentice-Hall, (1977).

    Google Scholar 

  • Thomas, T. and P. Lawler: Statistical Methods for Reliability Prediction. In: Electronic Materials Handbook, Vol. 1 Packaging, ASM International, 895 (1989).

    Google Scholar 

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Author information

Authors and Affiliations

  1. McPherson Reliability Consulting, LLC, Plano, TX, USA

    J. W. McPherson

Authors
  1. J. W. McPherson
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Problems

Problems

  1. 1.

    For certain implantable medical devices, the median time-to-failure is t50 = 87,600 h (10 years) and a logarithmic standard deviation of σ = 0.7. Assuming a lognormal distribution:

    1. (a)

      What is the instantaneous failure rate at 8 years?

    2. (b)

      What is the average failure rate after 8 years?

    3. (c)

      What fraction of the devices is expected to fail after 8 years?

    Answers:

    1. (a)

      λ @ t = 70,080 h = 1.24 × 10−5/h

    2. (b)

      <λ> after 780,080 h = 6.71 × 10−6/h

    3. (c)

      F = 0.375

  2. 2.

    A certain collection of capacitors has a Weibull time-to-failure distribution with a characteristic time-to-failure of t63 = 100,000 h and a Weibull slope of β = 1.2.

    1. (a)

      What is the instantaneous failure rate at 9 years?

    2. (b)

      What is the average failure rate after 9 years?

    3. (c)

      What fraction of the capacitors is expected to fail after 9 years?

    Answers:

    1. (a)

      λ @ t = 78,840 h = 1.14 × 10−5/h

    2. (b)

      <λ> after 78,840 h = 9.54 × 10−6/h

    3. (c)

      F = 0.528

  3. 3.

    For a mechanical component, fatigue data indicates that the median cycle-to-failure is (CTF)50 = 26,000 cycles and a logarithmic standard deviation of σ = 1.2. Assuming a lognormal distribution:

    1. (a)

      What is the instantaneous failure rate at 18,000 cycles?

    2. (b)

      What is the average failure rate after 18,000 cycles?

    3. (c)

      What fraction of the components is expected to fail after 18,000 cycles?

    Answers:

    1. (a)

      λ @ 18,000 cycles = 2.84 × 10−5/cycle

    2. (b)

      <λ> after 18,000 cycles = 2.65 × 10−5/cycle

    3. (c)

      F = 0.380

  4. 4.

    Certain mechanical components are found to corrode and can be described by a lognormal time-to-failure distribution with a characteristic time-to-failure of t50 = 50,500 h and a σ = 1.2.

    1. (a)

      What is the instantaneous failure rate at 40,000 h?

    2. (b)

      What is the average failure rate after 40,000 h?

    3. (c)

      What fraction of the components is expected to fail after 40,000 h?

    Answers:

    1. (a)

      λ @ t = 40,000 h = 1.41 × 10−5/h

    2. (b)

      <λ> after 40,000 h = 1.37 × 10−5/h

    3. (c)

      F = 0.423

  5. 5.

    Certain integrated circuits are found to fail, due to channel hot-carrier injection, and can be described by Weibull time-to-failure distribution with a characteristic time-to-failure of t63 = 75,000 h and a Weibull slope of β = 2.0.

    1. (a)

      What is the instantaneous failure rate at 60,000 h?

    2. (b)

      What is the average failure rate after 60,000 h?

    3. (c)

      What fraction of the circuits is expected to fail after 60,000 h?

    Answers:

    1. (a)

      λ @ t = 60,000 h = 2.13 × 10−5/h

    2. (b)

      <λ> after 60,000 h = 1.07 × 10−5/h

    3. (c)

      F = 0.473

  6. 6.

    Certain automobile tires are found to wear out according to a lognormal wearout distribution with characteristic parameters: (wear out)50 = 38,000 miles with a σ = 0.6.

    1. (a)

      What is the instantaneous wear-out rate at 32,000 miles?

    2. (b)

      What is the average wear-out rate after 32,000 miles?

    3. (c)

      What fraction of the tires is expected to wear out after 32,000 miles?

    Answers:

    1. (a)

      λ @ 32,000 miles = 3.25 × 10−5/mile

    2. (b)

      <λ> after 32,000 miles = 1.53 × 10−5/mile

    3. (c)

      F = 0.388

  7. 7.

    Certain hinges on doors are found to fail according to a Weibull distribution with the parameters: (number of closures)63 = 25,000 with a Weibull slope of β = 0.5.

    1. (a)

      What is the instantaneous failure rate at 18,000 closures?

    2. (b)

      What is the average failure rate after 18,000 closures?

    3. (c)

      What fraction of hinges is expected to fail after 18,000 closures?

    Answers:

    1. (a)

      λ @ 18,000 closures = 2.36 × 10−5/closure

    2. (b)

      <λ> after 18,000 closures = 4.71 × 10−5/closure

    3. (c)

      F = 0.572

  8. 8.

    Crowns, from a certain dental supply company, are found to fail according to a Weibull distribution with the parameters: t63 = 15.0 years and a Weibull slope of β = 1.0.

    1. (a)

      What is the instantaneous failure rate at 12 years?

    2. (b)

      What is the average failure rate after 12 years?

    3. (c)

      What fraction of crowns is expected to fail after 12 years?

    Answers:

    1. (a)

      λ @ t = 12 years = 6.67 × 10−2/year

    2. (b)

      <λ> after 12 years = 6.67 × 10−2/year

    3. (c)

      F = 0.551

  9. 9.

    Certain cell phones can start to fail after a number of drops. (Note: dropped phones not dropped calls.) The failures in a certain test are found to be described well by a Weibull distribution with the parameters: (number of drops)63 = 88 drops and Weibull slope of β = 0.6.

    1. (a)

      What is the instantaneous failure rate at 50 drops?

    2. (b)

      What is the average failure rate after 50 drops?

    3. (c)

      What fraction of phones is expected to fail after 50 drops?

    Answers:

    1. (a)

      λ @ 50 drops = 8.55 × 10−3/drop

    2. (b)

      <λ> after 50 drops = 1.42 × 10−2/drop

    3. (c)

      F = 0.510

  10. 10.

    Temperature-cycling of bi-metallic layers was found to produce delamination type failures conforming to a lognormal distribution with parameters: median cycle-to-failure of (cycles-to-failure)50 = 1,600 cycles and σ = 0.9.

    1. (a)

      What is the instantaneous failure rate at 1,300 cycles?

    2. (b)

      What is the average failure rate after 1,300 cycles?

    3. (c)

      What fraction of components is expected to fail after 1,300 cycles?

    Answers:

    1. (a)

      λ @ 1,300 cycles = 5.62 × 10−4/cycle

    2. (b)

      <λ> after 1,300 cycles = 4.04 × 10−4/cycle

    3. (c)

      F = 0.409

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McPherson, J.W. (2019). Failure Rate Modeling. In: Reliability Physics and Engineering. Springer, Cham. https://doi.org/10.1007/978-3-319-93683-3_8

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  • DOI: https://doi.org/10.1007/978-3-319-93683-3_8

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  • Publisher Name: Springer, Cham

  • Print ISBN: 978-3-319-93682-6

  • Online ISBN: 978-3-319-93683-3

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